Tutor profile: Vikas J.

Determine the area of the region enclosed by y = sinx, y=cosx, x=π/2, and the y-axis.

As we know at x = π/4 will be the intersection point for the two curves and also the position of the graph changes so we will find this integral in two parts : A = int. 0 to π/4 ( cosx - sinx ) + int. π/4 to π/2 ( sinx - cosx) = 0 to π/4 ( sinx + cosx ) + π/4 to π/2 ( -cosx - sinx ) = 2^(1/2) - 1 + 2^(1/2) - 1 = 2^(1/2) - 2 = 0.829

Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the given curves about the y-axis. y = e^x, y = e^-x, x = 1; about the y-axis

Volume = int. 2pix(e^x - e^-x) from 0 to 1 = 4pi/e

If x^3 + y^3 = 9 and x + y = 3, then the value of x^4+y^4 is

x^3+y^3 = (x + y) × (x^2 − xy + y^2) Putting given values of x^3+y^3 and (x + y) 9 = 3 × ((x+y)2 − 3xy) = 3 × (9 − 3xy) = 27 − 9xy 9xy = 18 xy = 2 x^4 + y^4 = (x^2 + y^2)2 - 2x^2y^2 = (x^2 + y^2)2 - 2*4 [Putting value of xy] = ((x + y)2 - 2xy)2 - 2*4 [Putting values of (x+y) and xy] = (9 - 4)2 - 2*4 = 17